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‎docs/0120-triangle.adoc‎

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[#0120-triangle]
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= 120. Triangle
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= 120. 三角形最小路径和
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{leetcode}/problems/triangle/[LeetCode - Triangle^]
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https://leetcode.cn/problems/triangle/[LeetCode - 120. 三角形最小路径和 ^]
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Given a triangle, find the minimum path sum from top to bottom. Each step you may move to adjacent numbers on the row below.
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给定一个三角形 `triangle` ，找出自顶向下的最小路径和。
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For example, given the following triangle
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每一步只能移动到下一行中相邻的结点上。*相邻的结点* 在这里指的是 *下标* 与 *上一层结点下标* 相同或者等于 *上一层结点下标 + 1* 的两个结点。也就是说，如果正位于当前行的下标 `i`，那么下一步可以移动到下一行的下标 `i` 或 `i + 1` 。
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[subs="verbatim,quotes,macros"]
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----
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[
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[*2*],
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[*3*,4],
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[6,*5*,7],
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[4,*1*,8,3]
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]
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----
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*示例 1：*
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The minimum path sum from top to bottom is `11` (i.e., *2* + *3* + *5* + *1* = 11).
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....
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输入：triangle = [[2],[3,4],[6,5,7],[4,1,8,3]]
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输出：11
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解释：如下面简图所示：
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2
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3 4
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6 5 7
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4 1 8 3
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自顶向下的最小路径和为 11（即，2 + 3 + 5 + 1 = 11）。
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....
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*Note:*
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*示例 2：*
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Bonus point if you are able to do this using only _O_(_n_) extra space, where _n_ is the total number of rows in the triangle.
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....
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输入：triangle = [[-10]]
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输出：-10
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....
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== 解题分析
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*提示：*
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image::images/0120-1.png[{image_attr}]
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* `+1 <= triangle.length <= 200+`
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* `triangle[0].length == 1`
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* `triangle[i].length == triangle[i - 1].length + 1`
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* `-10^4^ \<= triangle[i][j] \<= 10^4^`
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一图胜千言！
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*进阶：*
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自底向上计算逐层每个元素的最小和。最后即可得出最小路径和。
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* 你可以只使用 stem:[O(n)] 的额外空间（`n`为三角形的总行数）来解决这个问题吗？
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== 思考题
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尝试一下动态规划的思路。
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== 思路分析
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== 参考资料
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image::images/0120-1.png[{image_attr}]
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. https://leetcode-cn.com/problems/triangle/solution/120-san-jiao-xing-zui-xiao-lu-jing-he-by-alexer-66/[120. 三角形最小路径和 - 三角形最小路径和 - 力扣（LeetCode）^]
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. https://leetcode-cn.com/problems/triangle/solution/di-gui-ji-yi-hua-sou-suo-zai-dao-dp-by-crsm/[【Java】递归，记忆化搜索，再到DP - 三角形最小路径和 - 力扣（LeetCode）^]
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一图胜千言！
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自底向上计算逐层每个元素的最小和。最后即可得出最小路径和。
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[[src-0120]]
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[tabs]
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====
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一刷::
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+
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--
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[{java_src_attr}]
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----
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include::{sourcedir}/_0120_Triangle.java[tag=answer]
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----
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--
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二刷::
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+
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--
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[{java_src_attr}]
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----
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include::{sourcedir}/_0120_Triangle_2.java[tag=answer]
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----
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--
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====
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== 思考题
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尝试一下动态规划的思路。
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== 参考资料
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. https://leetcode.cn/problems/triangle/solutions/43600/120-san-jiao-xing-zui-xiao-lu-jing-he-by-alexer-66/[120. 三角形最小路径和^]
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. https://leetcode.cn/problems/triangle/solutions/23146/di-gui-ji-yi-hua-sou-suo-zai-dao-dp-by-crsm/[120. 三角形最小路径和 - 【Java】递归，记忆化搜索，再到DP^]
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‎logbook/202503.adoc‎

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|✅ 动态规划
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|{counter:codes2503}
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|{leetcode_base_url}/house-robber-iii/[337. House Robber III^]
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|{leetcode_base_url}/house-robber-iii/[337. 打家劫舍 III^]
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|{doc_base_url}/0337-house-robber-iii.adoc[题解]
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|✅ 动态规划+树 = 树形动态规划
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|{counter:codes2503}
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|{leetcode_base_url}/house-robber-iv/[2560. House Robber IV^]
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|{leetcode_base_url}/house-robber-iv/[2560. 打家劫舍 IV^]
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|{doc_base_url}/2560-house-robber-iv.adoc[题解]
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|❌ 看答案也一脸懵逼！
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|{counter:codes2503}
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|{leetcode_base_url}/delete-and-earn/[740. Delete and Earn^]
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|{leetcode_base_url}/delete-and-earn/[740. 删除并获得点数^]
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|{doc_base_url}/0740-delete-and-earn.adoc[题解]
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|❌ 把题目理解错了。处理后就是最简单的打家劫舍问题。
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|{counter:codes2503}
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|{leetcode_base_url}/unique-paths/[62. Unique Paths^]
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|{leetcode_base_url}/unique-paths/[62. 不同路径^]
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|{doc_base_url}/0062-unique-paths.adoc[题解]
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|✅ 动态规划
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|{counter:codes2503}
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|{leetcode_base_url}/unique-paths-ii/[63. Unique Paths II^]
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|{leetcode_base_url}/unique-paths-ii/[63. 不同路径 II^]
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|{doc_base_url}/0063-unique-paths-ii.adoc[题解]
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|✅ 动态规划
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|{counter:codes2503}
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|{leetcode_base_url}/unique-paths-iii/[980. Unique Paths III^]
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|{leetcode_base_url}/unique-paths-iii/[980. 不同路径 III^]
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|{doc_base_url}/0980-unique-paths-iii.adoc[题解]
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|❌ 明知是回溯，但在处理当前节点时，总把当前和下一步混在一起。有思路，没写出代码。
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|{counter:codes2503}
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|{leetcode_base_url}/minimum-path-sum/[64. Minimum Path Sum^]
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|{leetcode_base_url}/minimum-path-sum/[64. 最小路径和^]
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|{doc_base_url}/0064-minimum-path-sum.adoc[题解]
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|✅ 动态规划。直接在原有矩阵上操作即可。
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|{counter:codes2503}
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|{leetcode_base_url}/triangle/[120. 三角形最小路径和^]
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|{doc_base_url}/0120-triangle.adoc[题解]
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|✅ 动态规划。从底向上比从上向底效率更高。
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|===
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截止目前，本轮练习一共完成 {codes2503} 道题。

‎src/main/java/com/diguage/algo/leetcode/_0120_Triangle.java‎

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/**
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* Runtime: 1 ms, faster than 99.42% of Java online submissions for Triangle.
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* Memory Usage: 39.2 MB, less than 8.16% of Java online submissions for Triangle.
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*
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* @author D瓜哥 · https://www.diguage.com
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* @since 2020-02-07 23:14
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*/
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public int minimumTotal(List<List<Integer>> triangle) {
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int[] sums = new int[triangle.size() + 1];

‎src/main/java/com/diguage/algo/leetcode/_0120_Triangle_2.java‎

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package com.diguage.algo.leetcode;
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import java.util.Arrays;
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import java.util.List;
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public class _0120_Triangle_2 {
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// tag::answer[]
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/**
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* @author D瓜哥 · https://www.diguage.com
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* @since 2020-02-07 23:14
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*/
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public int minimumTotal(List<List<Integer>> triangle) {
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int[] dp = new int[triangle.size()];
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dp[0] = triangle.get(0).get(0);
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for (int i = 1; i < triangle.size(); i++) {
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List<Integer> row = triangle.get(i);
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// 注意：这里有个技巧，从后向前遍历，不会影响剩余元素的计算。
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// 如果从前向后遍历，则会影响下一个元素的计算。
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for (int j = row.size() - 1; j >= 0; j--) {
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if (j == 0) {
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dp[j] = row.get(j) + dp[j];
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} else if (j == row.size() - 1) {
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dp[j] = row.get(j) + dp[j - 1];
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} else {
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dp[j] = row.get(j) + Math.min(dp[j - 1], dp[j]);
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}
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}
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}
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return Arrays.stream(dp).min().getAsInt();
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}
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// end::answer[]
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public static void main(String[] args) {
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new _0120_Triangle_2()
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.minimumTotal(List.of(List.of(2), List.of(3, 4), List.of(6, 5, 7), List.of(4, 1, 8, 3)));
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}
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}

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